1. ## Continuous Functions!

Suppose h : (0,1)-> satisfies the following conditions:
for all x Э (0,1) there exists d>0 s.t. for all x' Э (x, x+d)n(0,1) we have h(x)<=h(x')

Prove that if h is continuous on (0,1) then h(x)<=h(y) whenever x,y Э (0,1) and x<=y. Use a counterexample to show that this results may not be true when h is continuous

I thought I could do this by using definitions, but its not working for me.

2. Originally Posted by emjones
Suppose h : (0,1)-> satisfies the following conditions:
for all x Э (0,1) there exists d>0 s.t. for all x' Э (x, x+d)n(0,1) we have h(x)<=h(x')

Prove that if h is continuous on (0,1) then h(x)<=h(y) whenever x,y Э (0,1) and x<=y. Use a counterexample to show that this results may not be true when h is continuous

I thought I could do this by using definitions, but its not working for me.

Suppose $h0,1)\to\mathbb{R}" alt="h0,1)\to\mathbb{R}" /> (presumably) satisfies the following conditions: for every $x\in(0,1)$ there exists some $\delta>0$ such that $x'\in (x,x+\delta)\implies h(x)\leqslant h(x')$. Prove that if $h$ is continuous then $h$ is non-decreasing. Give a counter example to show the reverse is isn
t true.
Right? What have you tried?

3. Hi drexel, sorry I don't know how to use the codes, but thank you for changing it for me.

I have tried to use a corollary in my notes relating to monotonic functions and differentiability, where the MVT is applied to [x,y]<[a,b] but im not getting anywhere.

Also I was asked to state the IVT earlier in the question, so thought this might be of use, but I don't think it can be used earlier! This is why im confused! Please help!

4. Originally Posted by emjones
Hi drexel, sorry I don't know how to use the codes, but thank you for changing it for me.

I have tried to use a corollary in my notes relating to monotonic functions and differentiability, where the MVT is applied to [x,y]<[a,b] but im not getting anywhere.

Also I was asked to state the IVT earlier in the question, so thought this might be of use, but I don't think it can be used earlier! This is why im confused! Please help!
You can't use the MVT! How do you know that $f$ is differentiable? How do you think you'd use the IVT, that's what I'm thinking too.